445 lines
12 KiB
Java
445 lines
12 KiB
Java
<!-- GFM-TOC -->
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* [1. 把数组中的 0 移到末尾](#1-把数组中的-0-移到末尾)
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* [2. 改变矩阵维度](#2-改变矩阵维度)
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* [3. 找出数组中最长的连续 1](#3-找出数组中最长的连续-1)
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* [4. 有序矩阵查找](#4-有序矩阵查找)
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* [5. 有序矩阵的 Kth Element](#5-有序矩阵的-kth-element)
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* [6. 一个数组元素在 [1, n] 之间，其中一个数被替换为另一个数，找出重复的数和丢失的数](#6-一个数组元素在-[1,-n]-之间，其中一个数被替换为另一个数，找出重复的数和丢失的数)
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* [7. 找出数组中重复的数，数组值在 [1, n] 之间](#7-找出数组中重复的数，数组值在-[1,-n]-之间)
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* [8. 数组相邻差值的个数](#8-数组相邻差值的个数)
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* [9. 数组的度](#9-数组的度)
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* [10. 对角元素相等的矩阵](#10-对角元素相等的矩阵)
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* [11. 嵌套数组](#11-嵌套数组)
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* [12. 分隔数组](#12-分隔数组)
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<!-- GFM-TOC -->
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# 1. 把数组中的 0 移到末尾
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[283. Move Zeroes (Easy)](https://leetcode.com/problems/move-zeroes/description/)
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```html
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For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].
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```
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```java
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public void moveZeroes(int[] nums) {
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int idx = 0;
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for (int num : nums) {
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if (num != 0) {
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nums[idx++] = num;
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}
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}
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while (idx < nums.length) {
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nums[idx++] = 0;
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}
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}
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```
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# 2. 改变矩阵维度
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[566. Reshape the Matrix (Easy)](https://leetcode.com/problems/reshape-the-matrix/description/)
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```html
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Input:
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nums =
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[[1,2],
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[3,4]]
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r = 1, c = 4
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Output:
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[[1,2,3,4]]
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Explanation:
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The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.
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```
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```java
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public int[][] matrixReshape(int[][] nums, int r, int c) {
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int m = nums.length, n = nums[0].length;
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if (m * n != r * c) {
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return nums;
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}
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int[][] reshapedNums = new int[r][c];
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int index = 0;
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for (int i = 0; i < r; i++) {
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for (int j = 0; j < c; j++) {
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reshapedNums[i][j] = nums[index / n][index % n];
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index++;
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}
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}
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return reshapedNums;
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}
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```
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# 3. 找出数组中最长的连续 1
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[485. Max Consecutive Ones (Easy)](https://leetcode.com/problems/max-consecutive-ones/description/)
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```java
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public int findMaxConsecutiveOnes(int[] nums) {
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int max = 0, cur = 0;
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for (int x : nums) {
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cur = x == 0 ? 0 : cur + 1;
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max = Math.max(max, cur);
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}
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return max;
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}
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```
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# 4. 有序矩阵查找
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[240. Search a 2D Matrix II (Medium)](https://leetcode.com/problems/search-a-2d-matrix-ii/description/)
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```html
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[
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[ 1, 5, 9],
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[10, 11, 13],
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[12, 13, 15]
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]
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```
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```java
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public boolean searchMatrix(int[][] matrix, int target) {
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if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
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int m = matrix.length, n = matrix[0].length;
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int row = 0, col = n - 1;
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while (row < m && col >= 0) {
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if (target == matrix[row][col]) return true;
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else if (target < matrix[row][col]) col--;
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else row++;
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}
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return false;
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}
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```
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# 5. 有序矩阵的 Kth Element
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[378. Kth Smallest Element in a Sorted Matrix ((Medium))](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/description/)
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```html
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matrix = [
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[ 1, 5, 9],
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[10, 11, 13],
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[12, 13, 15]
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],
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k = 8,
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return 13.
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```
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解题参考：[Share my thoughts and Clean Java Code](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/85173)
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二分查找解法：
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```java
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public int kthSmallest(int[][] matrix, int k) {
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int m = matrix.length, n = matrix[0].length;
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int lo = matrix[0][0], hi = matrix[m - 1][n - 1];
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while (lo <= hi) {
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int mid = lo + (hi - lo) / 2;
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int cnt = 0;
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n && matrix[i][j] <= mid; j++) {
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cnt++;
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}
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}
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if (cnt < k) lo = mid + 1;
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else hi = mid - 1;
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}
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return lo;
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}
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```
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堆解法：
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```java
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public int kthSmallest(int[][] matrix, int k) {
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int m = matrix.length, n = matrix[0].length;
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PriorityQueue<Tuple> pq = new PriorityQueue<Tuple>();
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for(int j = 0; j < n; j++) pq.offer(new Tuple(0, j, matrix[0][j]));
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for(int i = 0; i < k - 1; i++) { // 小根堆，去掉 k - 1 个堆顶元素，此时堆顶元素就是第 k 的数
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Tuple t = pq.poll();
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if(t.x == m - 1) continue;
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pq.offer(new Tuple(t.x + 1, t.y, matrix[t.x + 1][t.y]));
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}
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return pq.poll().val;
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}
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class Tuple implements Comparable<Tuple> {
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int x, y, val;
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public Tuple(int x, int y, int val) {
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this.x = x; this.y = y; this.val = val;
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}
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@Override
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public int compareTo(Tuple that) {
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return this.val - that.val;
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}
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}
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```
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# 6. 一个数组元素在 [1, n] 之间，其中一个数被替换为另一个数，找出重复的数和丢失的数
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[645. Set Mismatch (Easy)](https://leetcode.com/problems/set-mismatch/description/)
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```html
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Input: nums = [1,2,2,4]
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Output: [2,3]
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```
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```html
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Input: nums = [1,2,2,4]
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Output: [2,3]
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```
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最直接的方法是先对数组进行排序，这种方法时间复杂度为 O(NlogN)。本题可以以 O(N) 的时间复杂度、O(1) 空间复杂度来求解。
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主要思想是通过交换数组元素，使得数组上的元素在正确的位置上。
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```java
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public int[] findErrorNums(int[] nums) {
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for (int i = 0; i < nums.length; i++) {
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while (nums[i] != i + 1 && nums[nums[i] - 1] != nums[i]) {
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swap(nums, i, nums[i] - 1);
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}
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}
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for (int i = 0; i < nums.length; i++) {
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if (nums[i] != i + 1) {
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return new int[]{nums[i], i + 1};
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}
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}
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return null;
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}
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private void swap(int[] nums, int i, int j) {
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int tmp = nums[i];
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nums[i] = nums[j];
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nums[j] = tmp;
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}
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```
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类似题目：
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- [448. Find All Numbers Disappeared in an Array (Easy)](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/description/)，寻找所有丢失的元素
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- [442. Find All Duplicates in an Array (Medium)](https://leetcode.com/problems/find-all-duplicates-in-an-array/description/)，寻找所有重复的元素。
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# 7. 找出数组中重复的数，数组值在 [1, n] 之间
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[287. Find the Duplicate Number (Medium)](https://leetcode.com/problems/find-the-duplicate-number/description/)
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要求不能修改数组，也不能使用额外的空间。
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二分查找解法：
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```java
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public int findDuplicate(int[] nums) {
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int l = 1, h = nums.length - 1;
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while (l <= h) {
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int mid = l + (h - l) / 2;
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int cnt = 0;
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for (int i = 0; i < nums.length; i++) {
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if (nums[i] <= mid) cnt++;
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}
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if (cnt > mid) h = mid - 1;
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else l = mid + 1;
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}
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return l;
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}
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```
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双指针解法，类似于有环链表中找出环的入口：
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```java
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public int findDuplicate(int[] nums) {
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int slow = nums[0], fast = nums[nums[0]];
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while (slow != fast) {
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slow = nums[slow];
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fast = nums[nums[fast]];
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}
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fast = 0;
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while (slow != fast) {
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slow = nums[slow];
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fast = nums[fast];
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}
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return slow;
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}
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```
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# 8. 数组相邻差值的个数
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[667. Beautiful Arrangement II (Medium)](https://leetcode.com/problems/beautiful-arrangement-ii/description/)
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```html
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Input: n = 3, k = 2
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Output: [1, 3, 2]
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Explanation: The [1, 3, 2] has three different positive integers ranging from 1 to 3, and the [2, 1] has exactly 2 distinct integers: 1 and 2.
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```
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题目描述：数组元素为 1\~n 的整数，要求构建数组，使得相邻元素的差值不相同的个数为 k。
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让前 k+1 个元素构建出 k 个不相同的差值，序列为：1 k+1 2 k 3 k-1 ... k/2 k/2+1.
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```java
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public int[] constructArray(int n, int k) {
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int[] ret = new int[n];
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ret[0] = 1;
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for (int i = 1, interval = k; i <= k; i++, interval--) {
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ret[i] = i % 2 == 1 ? ret[i - 1] + interval : ret[i - 1] - interval;
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}
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for (int i = k + 1; i < n; i++) {
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ret[i] = i + 1;
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}
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return ret;
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}
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```
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# 9. 数组的度
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[697. Degree of an Array (Easy)](https://leetcode.com/problems/degree-of-an-array/description/)
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```html
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Input: [1,2,2,3,1,4,2]
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Output: 6
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```
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题目描述：数组的度定义为元素出现的最高频率，例如上面的数组度为 3。要求找到一个最小的子数组，这个子数组的度和原数组一样。
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```java
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public int findShortestSubArray(int[] nums) {
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Map<Integer, Integer> numsCnt = new HashMap<>();
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Map<Integer, Integer> numsLastIndex = new HashMap<>();
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Map<Integer, Integer> numsFirstIndex = new HashMap<>();
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for (int i = 0; i < nums.length; i++) {
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int num = nums[i];
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numsCnt.put(num, numsCnt.getOrDefault(num, 0) + 1);
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numsLastIndex.put(num, i);
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if (!numsFirstIndex.containsKey(num)) {
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numsFirstIndex.put(num, i);
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}
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}
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int maxCnt = 0;
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for (int num : nums) {
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maxCnt = Math.max(maxCnt, numsCnt.get(num));
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}
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int ret = nums.length;
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for (int i = 0; i < nums.length; i++) {
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int num = nums[i];
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int cnt = numsCnt.get(num);
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if (cnt != maxCnt) continue;
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ret = Math.min(ret, numsLastIndex.get(num) - numsFirstIndex.get(num) + 1);
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}
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return ret;
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}
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```
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# 10. 对角元素相等的矩阵
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[766. Toeplitz Matrix (Easy)](https://leetcode.com/problems/toeplitz-matrix/description/)
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```html
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1234
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5123
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9512
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In the above grid, the diagonals are "[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]", and in each diagonal all elements are the same, so the answer is True.
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```
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```java
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public boolean isToeplitzMatrix(int[][] matrix) {
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for (int i = 0; i < matrix[0].length; i++) {
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if (!check(matrix, matrix[0][i], 0, i)) {
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return false;
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}
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}
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for (int i = 0; i < matrix.length; i++) {
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if (!check(matrix, matrix[i][0], i, 0)) {
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return false;
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}
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}
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return true;
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}
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private boolean check(int[][] matrix, int expectValue, int row, int col) {
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if (row >= matrix.length || col >= matrix[0].length) {
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return true;
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}
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if (matrix[row][col] != expectValue) {
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return false;
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}
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return check(matrix, expectValue, row + 1, col + 1);
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}
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```
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# 11. 嵌套数组
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[565. Array Nesting (Medium)](https://leetcode.com/problems/array-nesting/description/)
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```html
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Input: A = [5,4,0,3,1,6,2]
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Output: 4
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Explanation:
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A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.
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One of the longest S[K]:
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S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}
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```
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题目描述：S[i] 表示一个集合，集合的第一个元素是 A[i]，第二个元素是 A[A[i]]，如此嵌套下去。求最大的 S[i]。
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```java
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public int arrayNesting(int[] nums) {
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int max = 0;
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for (int i = 0; i < nums.length; i++) {
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int cnt = 0;
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for (int j = i; nums[j] != -1; ) {
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cnt++;
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int t = nums[j];
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nums[j] = -1; // 标记该位置已经被访问
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j = t;
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}
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max = Math.max(max, cnt);
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}
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return max;
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}
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```
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# 12. 分隔数组
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[769. Max Chunks To Make Sorted (Medium)](https://leetcode.com/problems/max-chunks-to-make-sorted/description/)
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```html
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Input: arr = [1,0,2,3,4]
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Output: 4
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Explanation:
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We can split into two chunks, such as [1, 0], [2, 3, 4].
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However, splitting into [1, 0], [2], [3], [4] is the highest number of chunks possible.
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```
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题目描述：分隔数组，使得对每部分排序后数组就为有序。
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```java
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public int maxChunksToSorted(int[] arr) {
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if (arr == null) return 0;
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int ret = 0;
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int right = arr[0];
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for (int i = 0; i < arr.length; i++) {
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right = Math.max(right, arr[i]);
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if (right == i) ret++;
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}
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return ret;
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}
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```
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# 微信公众号
|
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更多精彩内容将发布在微信公众号 CyC2018 上，你也可以在公众号后台和我交流学习和求职相关的问题。另外，公众号提供了该项目的 PDF 等离线阅读版本，后台回复 "下载" 即可领取。公众号也提供了一份技术面试复习大纲，不仅系统整理了面试知识点，而且标注了各个知识点的重要程度，从而帮你理清多而杂的面试知识点，后台回复 "大纲" 即可领取。我基本是按照这个大纲来进行复习的，对我拿到了 BAT 头条等 Offer 起到很大的帮助。你们完全可以和我一样根据大纲上列的知识点来进行复习，就不用看很多不重要的内容，也可以知道哪些内容很重要从而多安排一些复习时间。
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